Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $k = \dfrac{q^2 + 3q - 4}{q - 1} \times \dfrac{3q + 3}{q + 4} $
Explanation: First factor the quadratic. $k = \dfrac{(q + 4)(q - 1)}{q - 1} \times \dfrac{3q + 3}{q + 4} $ Then factor out any other terms. $k = \dfrac{(q + 4)(q - 1)}{q - 1} \times \dfrac{3(q + 1)}{q + 4} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac{ (q + 4)(q - 1) \times 3(q + 1) } { (q - 1) \times (q + 4) } $ $k = \dfrac{ 3(q + 4)(q - 1)(q + 1)}{ (q - 1)(q + 4)} $ Notice that $(q - 1)$ and $(q + 4)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac{ 3\cancel{(q + 4)}(q - 1)(q + 1)}{ (q - 1)\cancel{(q + 4)}} $ We are dividing by $q + 4$ , so $q + 4 \neq 0$ Therefore, $q \neq -4$ $k = \dfrac{ 3\cancel{(q + 4)}\cancel{(q - 1)}(q + 1)}{ \cancel{(q - 1)}\cancel{(q + 4)}} $ We are dividing by $q - 1$ , so $q - 1 \neq 0$ Therefore, $q \neq 1$ $k = 3(q + 1) ; \space q \neq -4 ; \space q \neq 1 $